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# 数学代写|运筹学代写Operations Research代考|MATH2730 ECONOMIC INTERPRETATION OF THE DUAL

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## 数学代写|运筹学代写Operations Research代考|ECONOMIC INTERPRETATION OF THE DUAL

From the previous discussion we know that the objective function increases by 3 for a unit increase in the first resource. If we have to buy the resource we will be willing to pay a maximum of $₹ 3$ for the unit increase. Otherwise we will end up making a loss and it will not be profitable considering the purchase of the extra resource.
Result $3.8$
The value of the dual variable is the marginal value of the corresponding resource at the optimum.
Assume that the primal is a problem of a carpenter who makes tables and chairs. Now, the dual is the problem faced by the person who is assumed to be selling the resources to the carpenter. If the person sells the extra resource for a price less than $₹ 3$, the carpenter will buy and make more profit than what the problem allows him to make (which the seller would not want). On the other hand, if the seller charges more than $₹ 3$, the carpenter will not buy the resource and the seller cannot make money and profit. So both the carpenter and the seller will agree for $₹ 3$ (in a competitive environment) and each will make their money and the associated profit. The dual value is also called the shadow price of the resource at the optimum.
Consider Illustration $3.4$
Maximize $Z=3 X_1+4 X_2$
Subject to
\begin{aligned} X_1+X_2 & \leq 12 \ 2 X_1+3 X_2 & \leq 30 \ X_1+4 X_2 & \leq 36 \ X_1, X_2 & \geq 0 \end{aligned}
The optimal solution to the primal is:
$$X_1^=6, X_2^=6, u_3^=6, Z^=42$$
The optimal solution to the dual is:
$$Y_1^=1, Y_2^=1, W^*=42$$

## 数学代写|运筹学代写Operations Research代考|SIMPLEX METHOD SOLVES BOTH THE PRIMAL AND THE DUAL

Let us explain this using the formulation in Illustration 1.1. The simplex table is shown in the usual notation in Table 3.3.

In the optimal tableau, let us observe the $C_j-Z_j$ values. These are $0,0,-3$ and $-1$ for variables $X_1, X_2, u_1$ and $u_2$. We also know (from complimentary slackness conditions) that there is a relationship between $X_j$ and $v_j$ and between $u_j$ and $Y_j$.

The values of $C_j-Z_j$ corresponding to $X_j$ are the negatives of the values of dual slack variables $v_j$ and the values of $C_j-Z_j$ corresponding to $u_j$ are the negatives of the values of dual decision variables $Y_j$.
(We will see the algebraic explanation of this subsequently in this chapter.)
Therefore, $Y_1^=3, Y_2^=1, v_1^=0, \boldsymbol{v}_2^=0$. We also know from the complimentary slackness conditions that if $X_j$ is basic then $v_j=0$. This is also true because when $X_j$ is basic, $C_j-Z_j$ is zero. When $u_j$ is non-basic and has zero value, its $C_j-Z_j$ is negative at optimum indicating a nonnegative value of the basic variable $Y_j$.

The optimum solution to the dual can be read from the optimum tableau of the primal in the simplex algorithm. We need not solve the dual explicitly.

Let us look at an intermediate iteration (say with basic variable $u_1$ and $X_1$ ). The basic feasible solution is $u_1=1, X_1=4$ with non-basic variables $X_2=0$ and $u_2=0$. When we apply the above rule (and complimentary slackness conditions), we get the corresponding dual solution to be $Y_1=0, Y_2$ $=2, v_1=0, v_2=-1$ with $W=24$ (same value of $Z$ ).

This solution is infeasible to the dual because variable $v_2$ takes a negative value. This means that the second dual constraint $Y_1+2 Y_2 \geq 5$ is not feasible making $v_2=-1$. The value of $v_2$ is the extent of infeasibility of the dual, which is the rate at which the objective function can increase by entering the corresponding primal variable. A non-optimal basic feasible solution to the primal results in an infeasible dual when complimentary slackness conditions are applied. At the optimum, when complimentary slackness conditions are applied, the resultant solution is also feasible and hence optimal.